2.1 Mathematics and Metamathematics

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1 1 Version Information : More corrections (thanks to Peter Neumann) : Added material up to L = V = L and fixed more errors : Many thanks to Peter Neumann for the numerous little errors he spotted : I have rewritten and restructured some of the introductory material. At the moment these notes cover the course up to and including the definition of the cumulative hierarchy. 2 Background We work in first order logic with equality and are concerned with theories with one binary relation, written. The language is called the Language of Set Theory (LST). Our objects are going to be called sets and denoted (generally) by small latin letters. For quantifiers, we use the abbreviations and x y ψ x [x y ψ] x y ψ x [x y ψ]. Finally we single out a specific collection of formulae: we call a formula a 0 formula if every quantifier is bounded. Formally, we define the collection of 0 formula (in the metatheory) by recursion: it is the smallest collection of formulae that contains the atomic ones (x y and x = y) is closed under the logical connectives,,,, and if ψ is 0 then so are x y ψ and x y ψ (y must be a variable). Note that it might not make sense to talk about the collection of 0 formulae. This depends on your metatheory! But for any specific formula, it is trivial to verify (in any reasonable metatheory) whether or not this is Mathematics and Metamathematics Note that we are, mostly, studying ɛ-structures from the outside. Of course, the question arises in which theory we do this, i.e. what our metatheory is. We are purposely vague about this. Any reasonable finitistic metatheory should work. It is however important that we do not confuse the metatheory and the theory. We will, for example, prove theorems within our theory (see for example the section on Ordinals). We will also prove theorems which cannot even be stated about in our theory: for example if we say for every formula φ of LST..., this is a theorem in the metatheory. 1

2 3 Defined Notions 3.1 Classes A class C φ is a formula φ(t) with one free variable (here t). Instead of writing φ(x) we write x C φ. We frequently do not give an explicit formula for a class but simply denote it by a capital latin letter. In the following we define various notions for sets. When these make sense for classes, we will use the same notation. As an example of this, if C φ is a class given by the formula φ(t) we write The Universe x C φ ψ x [x C φ ψ] x [φ(x) ψ]. We will write U for the class given by the formula t = t (so that trivially x x U). Note however that the formula x U ψ is not 0 : it is after all just an abbreviation for the formula x [x = x ψ] and this is not Defined Notions To enhance readability of our formulae, we define abbreviations for formulae of LST. The formulae given are somewhat arbitrary, there are plenty others which are equivalent. We try to give 0 -formulae - this is important later on when we talk about absoluteness. We will often define something like z = {x, y} φ(x, y, z) but then use {x, y} w to mean t w [t = {x, y}] or more explicitly t w φ(x, y, t). This is 0 provided that the definition of w = {x, y}, namely φ, was 0. We will also write w {x, y}. This is harder to define abstractly, so we are (sometimes) explicit about it. 3.3 Comprehension For a formula φ z = {y : φ(y)} t [t z φ(t)]. 3.4 Subset x y t x t y. 2

3 3.5 Emptyset z = t z [t t] t t t t z t z =. 3.6 Unordered Pair z = {x, y} x z y z t z [t = x t = y] and We use the shorthand t {x, y} t = x t = y. {x} {x, x}. 3.7 Ordered Pair z = x, y z = {{x}, {x, y}} t z [t = {x, x}] t z [t = {x, y}] t z [t = {x, x} t = {x, y}]. Note that the first statement only makes sense if {x} and {x, y} are objects of our model, whereas the full formula is defined independent of the existence of {x} and {x, y}. z is an ordered pair a z b z x a y b z = x, y We will frequently refer to the first and second coordinate of the ordered pair. To do so we define and x = π 1 (z) z is an ordered pair b z y b [z = {x, y}] y = π 2 (z) z is an ordered pair a z x a [z = x, y ]. We will also use the fact that z, z are ordered pairs x, y [π 1 (z) = x = π 1 (z ) π 2 (z) = y = π 2 (z )] 3.8 Relations We are only interested in binary relations, so simply use the word relation for binary relations. r is a relation t r t is an ordered pair. 3

4 r is a transitive relation r is a relation u r v r [π 2 (u) = π 1 (v) π 1 (u), π 2 (v) z]. It is somewhat unclear how the above is translated into a formula of LST (i.e. how to eliminate the defined notions). Here is one way to do that: r is a transitive relation r is a relation u r v r x, y, z [[x = π 1 (u) y = π 2 (u) y = π 1 (v) z = π 2 (v)] w r [w = {x, z}]]. In this formula, not every quantifier is bounded, but we can write one down where every quantifier is indeed bounded: replace the last bit by a u x a b u y b c v z c [u = x, y v = y, z w r [w = x, z ]]. The more you do this, the more unwieldy the formula becomes. We will in the future hence avoid these complicated formulae, but you should always check that you can carry out this replacement. We will need to consider relations on sets so we will define and as well as the classes (!) x dom(r) r is a relation z r [x = π 1 (z)] y ran(r) r is a relation z r [y = π 2 (z)] dom(r) = {x : x dom(r)} ran(r) = {y : y ran(r)}. Although these definitions seem self-referential, they are not: the string x dom(r) is simply replaced by r is a relation z r [x = π 1 (z)]. In particular, if r is not a relation then dom(r) =. Note that we seem to have defined infinitely many classes here, which is of course a bad thing. But in practice we will only ever use finitely many instances of this or insist on r being a set and having sufficiently many axioms available to show that this implies that dom(r) and ran(r) are sets. So we can define and r is a relation on x r is a relation dom(r) x ran(r) x r is a transitive relation on x r is a relation on x u, v, w x [ u, v r v, w r u, w r]. Note that this is 0 (provided r and x are sets). Similarly r is a reflexive relation on x r is a relation on x t x [ t, t r] r is an irreflexive relation on x r is a relation on x t x [ t, t r]. Of course, usually we write binary relations in infix notation, i.e. arb instead of a, b r. 4

5 3.9 Orders r is a partial strict order on x r is a transitive, irreflexive relation on x. r is a total strict order on x r is a transitive, irreflexive relation on x u, v x [ u, v r v, u r u = v]. r is a strict well-order on x r is a strict total order on x p [p x p m p [ t p [t r m]]] Note that this latter definition is not a 0 -formula (and in fact cannot be replaced by one) Functions f is a function f is a relation x dom(f) y ran(f) y ran(f) [xfy xfy y = y ] We can replace this by a 0 formula (as it stands it is not since dom(f) and ran(f) are defined notions and not variables). If we have shown that f is a function and x dom(f), we will write f(x) for the unique y such that xfy Class functions Note that the notion of being a function makes sense for classes. In this case we talk about class functions. We spell out what this means formally (because class functions are central to what we will be doing). Definition 1. Suppose φ(t) is a formula of LST. We say that φ is a class function if and only if t [φ(t) t is an ordered pair] x y y [φ( x, y ) φ( x, y ) y = y ]. Usually we denote class functions by capital latin letters and write F (x) = y for F ( x, y ). If A, B are classes, we say that F : A B is a class function if and only if 3.12 Union F is a class function dom(f ) = x ran(f ) B z = x [ y x t y [t z]] [ t z y x [t y y x]] and z = x y x z y z t z [t x t y]. 5

6 3.13 Powerset z = P (x) t [t z t x] Successor z = x + 1 z = x {x, x} x z x z t z [t x t = x] Inductive Set Ind(x) x t x [t + 1 x]. Axiom Summary Recall the definitions: x y t x t y z = t z t t z = {x, y} x z y z t z [t = x t = y] z = x t z y x [t y] y x t y [t z] z = P (x) t [t x t z] t z [t x] z = S(x) x z t x [t z] t z [t = x t x] and then state (any free variables are implicitly universally quantified) Extensionality x y y x x = y Separation For each formula φ(v 1,..., v n, v n+1 ) with all free variables shown x y y = {z x : φ (a 1,..., a n, z)}. Emptyset Pairing Union Powerset z z = x y z z = {x, y} x y y = x x y y = P (x) 6

7 Replacement For each formula φ(p 1,..., p n, t, u) (with all free variables displayed) a 1,..., a n d x d!y φ(a 1,..., a n, x, y) z z = {y : x d φ(a 1,..., a n, x, y)} Infinity Foundation z [ x z x = y z w z w = S(y)] x [x y x [ z x [z y]]] Choice We will eventually state the Well-ordering Principle which is equivalent to the usual Axiom of Choice (see Part b, Set Theory) Axiom Systems We will sometimes work in axiom systems which do not include all of the above axioms. Standard abbreviations are ZF (sometimes denoted by ZF ): Extensionality+Separation+Emptyset+ Pairing + Union + Powerset + Replacement + Infinity. ZF: ZF + Foundation. ZFC: ZF + Choice. 4 Alternative Axiom Summary If you would like to remove the defined notions above, you may end up with something like the following: Extensionality x y [ z [z x z y] x = y] Separation For each formula φ(p 1,..., p n, t) (with all free variables displayed) Emptyset a 1,..., a n x y z [z y z x φ (a 1,..., a n, z)] This is also called the Comprehension Scheme and is often written as a 1,..., a n x y y = {z x : φ (a 1,..., a n, z)}. x y y x Note that Emptyset in fact follows form Separation with the formula φ(t) t t and the existence of any x (or by assuming that all models are non-empty). 7

8 Pairing x y z t [t z [t = x t = y]] Note that this definition of z = {x, y} t [t z [t = x t = y]] is not 0, so we would have to prove manually that it is absolute for transitive non-empty classes A B (assuming enough of ZF ). With Comprehension this is equivalent to x y z [x z y z] Union x y w [w y z [w z z y]] With Comprehension this is equivalent to x y z w [w z z x w y] Powerset x y z [z y w [w z w x]] With Comprehension this is equivalent to x y z [ w [w z w x] z y] Replacement For each formula φ(p 1,..., p n, t, u) (with all free variables displayed) a 1,..., a n d x d!y φ(a 1,..., a n, x, y) z z = {y : x d φ(a 1,..., a n, x, y)} Infinity Foundation [ z x z x = y z w z w = ] {y, {y, y}} x [x y x [ z x [z y]]] Choice We will eventually state the Well-ordering Principle which is equivalent to the usual Axiom of Choice (see Part b, Set Theory). 8

9 5 Relativization and Absoluteness I will only define relativization and absoluteness for LST, but it can easily be defined for any theory. The examples in lectures are intended to be given in the relevant theories. Definition 2. Given a class A and a formula φ of LST, the relativization of φ to A is the formula φ A where each quantifier is bounded by A. Formally, by induction on the complexity of the formula we define (x = y) A (x = y); (x y) A (x y); ( φ) A φ A ; (φ ψ) A (φ A ψ A ); (φ ψ) A (φ A ψ A ); (φ ψ) A (φ A ψ A ); (φ ψ) A (φ A ψ A ); ( x φ) A x A φ A x (x A φ A ); ( x φ) A x A φ A x (x A φ A ). If a 1,..., a n A and φ free variables x 1,..., x n then we also write A = φ(a 1,..., a n ) φ(a 1,..., a n ) A. This is just your standard interpretation of φ in the model A, except that of course A might not be a model (it might not be a set but only a class) and the above is purely syntactic. Remark 1 (Expanding on the previous sentence). Intuitively, the relativization of a formula φ to a class A is simply the interpretation of of φ in the model (A, ). The problem we are facing is that if A is really a class and not a set (in our meta-theory), then it does not make sense to talk about the model (A, ). We cannot (or do not want to) use the semantic notions, so have to rely on purely syntactical definions. If you are willing to work in a meta-theory in which you can show that if a theory is consitent, then it has a model this problem goes away. You will assume that ZF is consistent, and call its model (U, ) (so U really does exist as an object of study - just like you usually assume in ordinary mathematics that R really does exist as an object of study). You then work with a subobject A of U and can interpret formulae in (A, ) in the classical model theoretic sense. If you do this, formally we will then be showing that if ZF is consistent, then so is ZF and ZFC and ZFC + CH. But of course, it is very likely that if 9

10 your meta-theory is strong enough to show that if a theory is consistent, then it has a model, then it will include some version of Choice, so that somewhat defeats the purpose. If you are prepared to formally jump through the extra logical hoops, then you can get away with a much weaker meta-theory (some finitistic meta-theory) and prove the stronger statement if ZFC + CH is inconsistent, then so is ZF. For (most) practical purposes it is enough to think about the intuitive definition, but keep these comments in the back of your mind. Definition 3. Suppose A is a class and φ(x 1,..., x n ) is a formula of LST (with all free variables shown) and Γ is a collection of sentences. We say that A models φ (or A believes φ) in the context Γ, written A = Γ φ if and only if Γ a 1,..., a n A φ(a 1,..., a n ) A. If is a collection of formulae of LST, we say that A models (or A believes in ) in the context Γ, written A = Γ if and only if for each φ from, A = Γ φ. Usually we do not specify Γ explicitly and take it to be some suitable subcollection of ZFC. Again, intutively this says that the two models (A, ) and (B, ) have the same believe about some formula φ whenever that makes sense (i.e. all free variables are instatiated with elements of both A and B). Since formally, we will not be working with models (of ZF or ZF or ZFC), we have to give a syntactic definition which only relies on the existence of proofs (finite sequences of finite strings of symbols which follow some easily checkable rules), i.e. a finitistic meta-theory. Definition 4. Suppose φ(x 1,..., x n ) is a formula of LST (with all free variables shown), A, B are classes and Γ is a collection of sentences such that Γ A B. We say that φ is absolute for A, B in the context Γ if and only if Γ a 1,..., a n A (φ(a 1,..., a n ) A φ(a 1,..., a n ) B ). Note that often we do not mention Γ. For us, it will always be a subset of ZFC and we take it whatever subset we need in a proof. Recall that a class A is transitive if and only if x A x A. Definition 5. Suppose φ is a formula of LST. We say that φ is absolute (for transitive classes satisfying ) [in the context Γ] if and only if for any transitive classes A, B such that Γ A = and Γ B = ) and Γ A B, φ is absolute for A, B in the context Γ. Lemma 1. 0 formulae are absolute for transitive classes. Proof. Let A, B be transitive classes such that A B. We do this by induction of the complexity of the formula. By the definition of relativization, the only interesting steps are the quantifiers. We do the existential case. The universal case is similar or can be deduced by the replacement. 10

11 Case φ x y ψ(x, y, x 1,..., x n ): Fix x 1,..., x n, y A. First assume A = x y ψ(x, y, x 1,..., x n ), i.e. x A (x y ψ(x, y, x 1,..., x n ) A ). Find x A such that x y ψ(x, y, x 1,..., x n ) A. Since A B we have x B and by absoluteness of ψ (inductive assumption) ψ(x, y, x 1,..., x n ) B. Thus B = x y ψ(x, y, x 1,..., x n ). Conversely, assume that B = x y ψ(x, y, x 1,..., x n ), i.e. x B (x y ψ(x, y, x 1, dots, x n ) B ). Find x B such that x y A and ψ(x, y, x 1,..., x n ) B. By transitivity of A we have x A and by absoluteness of ψ (inductive assumption) ψ(x, y, x 1,..., x n ) A. Thus A = x y ψ(x, y, x 1,..., x n ). Remark 2. Note that the above in fact shows that if ψ(x) is absolute for A B and A = x ψ then B = x ψ. It is only for the reverse direction that we need transitivity and the fact that the quantifier is bounded. Remark 3. This is a classic meta-theorem. View it as a factory that produces proofs: we will be interested (mostly) in specific applications of this to specific formulae. So for example we will have reason to show that pairing is absolute, i.e. that if (we can prove that) A B are non-empty transitive classes, then we can prove that x, y, z A [A = z = {x, y} B = z = {x, y}] (this is a sentence in LST). We could (and you should) go through these proofs by hand, the above lemma is simply an abbreviating step. 6 The Ordinals In this section we develop just enough of the theory of ordinals to prove the results of the next section. You are strongly encouraged to consult the literature for more results about ordinals. Unless otherwise specified, we work in ZF Powerset (it may be fun to figure out exactly which axioms are needed to prove the results below). Definition 6. On = {α : α is transitive is a well-order on α} For the benefit of the reader we write out α On: x α y x y α x α [x x] x, y, z α [x y y z x z] [α is transitive] x, y α [x y y x x = y] [ is a strict total order on α] x [x α x m x z m z x] [ is well-founded on α] Lemma 2. ZF α On α is transitive and totally ordered by Thus if A, B are non-empty transitive classes satisfying (enough of) ZF then x On is absolute. 11

12 Lemma α On α α; 2. On; 3. α On α + 1 := α {α} On; 4. α, β On α β On; 5. α, β On [α β α β α = β]; 6. α On α On; Proof. 1. Suppose α On and α α. Then by irreflexivity, α α, a contradiction. 2. is vacuously transitive and is vacuously a well-order on. 3. Suppose α On. α + 1 is transitive: Suppose x y α + 1. Either y α and then by transitivity of α, x α or y = α and then x α. In any case x α α+1. is irreflexive on α + 1: Suppose x α + 1. Then either x α so that by assumption x x or x = α and thus α α by (1). is transitive on α + 1: Suppose x, y, z α + 1 and x y z. If x, y, z α then we get x z from the assumption that α On. Next, assume x = α. Then y α by (1) so y α. By transitivity of α we have α = x α contradicting (1). Similarly if y = α (use z α). So, suppose z = α. Then by transitivity of α we get x α = z. is a total order on α + 1: We need to verify trichotomy, so suppose x, y α + 1. If x = y = α we are done. If x, y α we are done by the assumption that α On. If (wlog) x α, y = α then we are also done. is well-founded on α + 1: Now suppose = x α + 1. If x = {α} then α is clearly the -minimal element of x (as α α by (1)). So assume x α : then x α α so x α has an -minimal element m. Now if t x then either t α so that t m by construction of m or t = α. But α = t x α gives α α by transitivity of α, a contradiction. Hence m is -minimal in x. 4. Suppose α, β On and x y α β. By transitivity of α and β respectively we obtain that x α β. Now observe that the restriction of any well-founded strict total order to a subset (or subclass) is a wellfounded strict total order. 12

13 5. Suppose α, β On with α β but α β. Then β \ α is a non-empty subset of β so has an -minimal element x. We claim x = α: Firstly if t x but t α then by transitivity of β we have t β \ α contradicting -minimality of x. Thus x α. Secondly, if t α then by assumption (namely α β) t β. So one of x t or t x or t = x must be true. If x t then by transitivity of α we have x α contradicting x β \ α. Similarly if t = x. Thus t x giving α x, as required. 6. Suppose α On and x α. If r t x then r x by transitivity of on α. Also, because α is transitive we have x α so that restricts to a well-founded strict total order on x. Hence x On. Theorem 1. is a well-founded strict total order on On. Theorem 2. If X is a non-empty subclass of On, then X has an -minimal element. Proof. We show: is a strict total order on On: By the lemma, is irreflexive. If α, β, γ On and α β γ then by transitivity of γ we have α γ as required. For totality, assume that α, β On. Let x = α β and note that by the lemma x On and x α and x β. So by the lemma (x α or x = α) and (x β or x = β). Unless x α and x β we have one of α β, β α or α = β. But if we assume x α and x β. Then x α β = x contradicting the lemma. is well-founded on On: Suppose that x is a non-empty subset of On. Pick α x and let y = α x. If y = then α is -minimal in x. If on the other hand y then y is a non-empty subset of α so has an -minimal element m. By construction m x. If t x m then as m α and α is transitive we have t α so t x α = y contradicting minimality of y. Thus x m = and m is in fact -minimal in x. Note that this works independent of whether x is a class or a set. Remark 4. Again, note that the first theorem is in fact a theorem of ZF Powerset. The second theorem is a result in our meta-theory. It cannot even be stated in the theory, since we don t have classes in our theory. Corollary 1 (Induction on On - another meta-theorem). Suppose that φ(t) is a formula. Then ZF Powerset [ α On [ β α φ(β)] φ(α)] α On φ(α). 13

14 Definition 7. Suppose that α On. α is a successor β On α = β + 1 α is a limit α α is not a successor α is finite [α is a successor α = ] t α [t = t is a successor] x = ω x On Ind(x) t x [t is finite]. Proof. We need to check that x = ω is in fact a definition, i.e. that if x and y satsify the RHS then x = y: Note that x, y On gives wlog that x y (or x = y and we are done). Then x is finite so in particular a successor, say β +1. But then β x and as x is inductive, x = β + 1 x, a contradiction. Lemma 4. z [Ind(z) ω z] Proof. Suppose z is an inductive set. By induction on the elements of ω we show ω z. Formally, suppose n ω \ z. As is well-founded on ω, we may assume that n is -minimal in ω \ z. As Ind(z) we cannot have n =. Thus n = m + 1 for some m and by transitivity of ω we have m ω. By minimality of n, we must have m z. But Ind(z) then gives n = m + 1 z, a contradiction. Lemma 5. Suppose x On and x is a set. Then x On and sup x = x. Proof. If r t x then there is α x with r t α so r α x since α is transitive. Next, if t x then t α for some α x to t On. Hence x On and so x is well-ordered by. Now let α 0 = x. If β x and t β then t α 0 so that β α 0. Thus α 0 is an upper bound for x. Finally, if α On is an upper bound for x and t α 0 then find β x with t β α (since α is an upper bound for x). Thus α 0 α as required. 7 Recursion A few more meta-theorems. Theorem 3 (The informal Recursion Theorem). Suppose F is a class function on U and that a U. Then there is a unique class function G on On such that: 1. G(0) = a; 2. α On G(α + 1) = F (G(α)); 3. γ Lim G(γ) = {G(α) : α < γ} We restate this more formally: 14

15 Definition 8. Suppose that F is a class function on U and that a U. We let ψ F,a (α, g) α On g is a function on α + 1 g(0) = a β α [g(β + 1) = F (g(β))] [ γ Lim α + 1 g(γ) = ] {g(β) : β γ} expressing that g is a function on α + 1 and that g satisfies the conditions above on its domain. We let G F,a { α, y : α On [ g [ψ F,a (α, g) α, y g]]}. Theorem 4 (The Recursion Theorem). If ZF Powerset proves that F is a class function on U and a U, then ZF Powerset proves that G F,a is a class function on On and G F,a (0) = a β On G F,a (β + 1) = F (G F,a (β)) γ Lim G(γ) = {G(β) : β γ} and ZF Powerset proves that if G, H are class functions satisfying the displayed formula then α On H(α) = G(α). Even this is (technically speaking) not formal enough. For once, I give the formal version of this theorem without the uniqueness bit (with defined notions not yet eliminated): Theorem 5 (The Formal Recursion Theorem). Suppose φ(t) is a formula with free variable displayed. As always we write F = {x : φ(x)} for the class defined by φ ZF Powerset [F is a class function on U] a U G F,a is a class function on On G F,a (0) = a β On G F,a (β + 1) = F (G F,a (β)) γ Lim G(γ) = {G(β) : β γ} As an exercise, you can try to remove all the defined notions from this theorem. Needless to say, for examination purposes the Informal Recursion Theorem is sufficient (though for the proof you will likely have to define ψ F,a and G F,a ). 15

16 Lemma 6. Suppose ZF Powerset proves that F is a class function on U and that a U. ZF Powerset proves that for every α On there is a unique g such that ψ F,a (α, g). Proof. Induction on On: For α = 0, use g = { 0, a } (which exists by Pairing) and uniqueness follows from Extensionality. For α = β + 1: Let g be the unique function for β which exists by the inductive hypothesis. Let g = g { α, g (β) } (this is a set by Union, Pairing and inductive hypothesis). It is clear that g is a function on α + 1 which satsifies the conditions (noting that α+1\β +1 = {α} so there are no new limit ordinals). Uniqueness follwos once more by the requirement of the conditions and Extensionality. For γ Lim: For each β γ, let g β be the unique function given by the lemma for β and note that if β < β then ĝ = g β β+1 is a function on β + 1 that satisfies ψ F,a (β, ĝ), so must equal g β. Next note that {g β : β γ} = {g : β γ ψ F,a (β, g)} is a set by Replacement. So g = { γ, {g β (β) : β γ} } {gβ : β γ} is a set by Union and Pairing. Now, by the note (that the g β agree on the common elements of their domains) g is a function on γ + 1 and it is clear that g satisfies the conditions. Proof of the Recursion Theorem. Note that we do two proofs at once (for efficiency reasons): one proof giving the theorem as stated and one giving the theorem up to some limit ordinal γ. The latter only uses the previous lemma for α < γ, whereas the former uses the lemma for all ordinals. We start by demonstrating that G is indeed a class function on On (resp. γ): it is clear from the definition of G that G is a relation from On to U. For α On (resp. α < γ), we apply the previous lemma to see that there is g such that ψ F,a (α, g). Since α α + 1, there is y U with α, y g. Hence the domain of G is all of On. Finally assume that there is α (resp. α < γ) and y, y U with α, y, α, y G. Then we may take the witnessing g, g and note that by the previous lemma g = g so that y = y as required. Finally we need to check that G satisfies the formula (up to γ). But this follows directly from the definition of G and induction on On. Remark 5. Let us note that if A, B are transitive non-empty classes satisfying (enough of) ZF, On A B, F is absolute for A, B and a A is given by a defined notion absolute for A, B then G is absolute for A, B. This requires some careful checking (e.g. ψ F,a is absolute for A, B) but is essentially straightforward once we have that On is absolute for A, B. There are some variations on recursion which we will use: Theorem 6. Suppose A, B are classes and that F : A B B and H : A B are class functions. Then there is a unique class function G : A On B such that: G(x, 0) = H(x); 16

17 α On G(x, α + 1) = F (x, G(x, α)); γ Lim G(γ) = β<γ G(β). Proof. As the proof of the Recursion Theorem, with ψ now being ψ F,H (x, α, g) where g(0) = a is replaced by g(0) = H(x) and G F,H { x, α, y : α On x A [ g [ψ F,H (x, α, g) α, y g]]}. Also, we may sometimes only define G up to some fixed ordinal α 0 (typically ω). The proof works as before, except that we insist on α + 1 < α 0 throughout. Theorem 7. Suppose A, B are classes and that F : A B B and H : A B are class functions and 0 < α 0 On. Then there is a unique class function G : A α 0 B such that: G(x, 0) = H(x); α On [α + 1 α 0 G(x, α + 1) = F (x, G(x, α))]; γ Lim α 0 G(γ) = β<γ G(β). If α 0 = ω then Replacement is not needed. If A is a set (and Replacement holds or A is a singleton (finite set)) then G is a set. 8 The Cumulative (von Neumann) Hierarchy V Definition 9. Let F (x) = P (x). Then ZF proves that F is a class function and we apply the Recursion Theorem with a = to obtain a class function V on On such that (writing V α for V (α)) V 0 = ; α On V α+1 = P (V α ); γ Lim V γ = β<γ V β. Abusing notation, we write V = α On V α. V is called the Cumulative (von Neumann) Hierarchy. Lemma 7. ZF proves that for all α On: V α is transitive; V α V α+1 ; α V α+1 ; Hence V is a transitive non-empty class containing On and for α, β On we have α β V α V β. Proof. We prove this by simultaneous induction on α: 17

18 Base Case: α = : Vacuously V 0 = is transitive and contained in V 1 = P ( ) = { }. By inspection 0 = V 1. Successor Step: hold for α + 1: Suppose (1)-(3) hold for α On. We will show that they 1. Let r t V α+1 = P (V α ). Then t V α (by construction) so r V α V α+1 (by inductive hypothesis) as required. 2. Let t V α+1. Then t V α V α+1 (by inductive hypothesis) so that t P (V α+1 ) = V α By inductive hyothesis α V α V α+1 and α V α+1 so that α+1 V α+1 giving α + 1 P (V α+1 ) = V α+1+1. Limit Step: Suppose γ Lim and (1)-(3) hold for α < γ. Recall that V γ = α<γ V α. 1. If r t V γ then find α < γ with r t V α and by transitivity of V α we have r V α V γ. 2. Suppose t V γ. Then t V α for some α < γ so by transitivity of V α, t V α V γ, giving t P (V γ ) = V γ For each α < γ we have α + 1 < γ (since γ is not a successor) and thus α V α+1 V γ. Hence γ V γ so γ V γ+1. The Hence now follows from (3) and induction. We prove a little utitlity lemma (we only need this to show that V = Replacement but it makes the other axioms easier to check). Lemma 8. ZF proves that if z is a set and z V then z V. Proof. For t z let α t be the least ordinal with t V α (formally, we write down a formula expressing this and check that this is a class function on z). Then β = sup {α t + 1 : t z} is a set by Replacement (and some others which we needed to establish that sets of ordinals have sups). By the previous lemma z V β so z V β+1 as required. Theorem 8. For every φ ZF, V = φ, i.e. ZF φ V. In fact, if A is a transitive, non-empty class such that A = Separation and x A z A x z then A = ZF. Proof. We prove this for V but remark that the proof also works for the A as specified. Extensionality: Extensionality follows from transitivity of V. 18

19 Separation: We do Separation next (since then the weaker versions of the axioms stated in here imply the stronger versions with in place of ): so let φ(t; v 1,..., v n ) be a formula of LST with all free variables shown and let a 1,..., a n, y V. Define z = { t : t V t y φ(t; a 1,..., a n ) V }. Then z is a set by Separation (in U) and by the little lemma z V (in fact, if y V α then z V α so z V α+1 avoids the lemma and hence the use of Replacement). We need to check that V = t [t z t y φ(t; a 1,..., a n )] or more explicitly (using the defining of =) that ZF t V [ t z t y φ(t; a 1,..., a n ) V ]. So let t V and assume t z. By definition of z (in U) t y φ(t; a 1,..., a n ) V as required. Now let t V and assume t y φ(t; a 1,..., a n ) V. Then by definition of z (in U) we have t z as required. Replacement: We follow the strategy from above: given a formula φ we cook up a formula ψ and apply Replacement with ψ in U to obtain some z. Then we check z V and that V believes the right stuff about z. Since the format of the axiom is a little more complicated, the proof is slightly messier: So suppose that φ(x, t, d; v 1,..., v n ) is a formula of LST with all free variables shown. Let a 1,..., a n, d V and assume that V = x d!yφ(x, y, d). Since V is transitive and d V this gives ZF x d!y V φ(x, y, d) V. So let ψ(x, y, d) be y V φ(x, y, d; a 1,..., a n ) V and observe that we may apply Replacement in U to obtain c such that x d y cψ(x, y, d). Let ĉ = c V V (we are using Separation here - this is unnecessary if we had used the stronger form of Replacement) so ĉ V. Now we show that V = x d y ĉφ(x, y, d; a 1,..., a n ). Assume x d and obtain y c (not ĉ!) such that ψ(x, y, d). But then y V, so y ĉ and φ(x, y, d, a 1,..., a n ) V. Pairing: The formula defining x = {y, z} is 0 so absolute for V, U and if y, z V then {y, z} V so {y, z} V. (Again, strictly speaking we don t need the utitility lemma: we can choose α, β so that y V α, z V β and wlog α β. Then {y, z} V β so {y, z} V β+1.) 19

20 Union: Write it out in detail as an exercise. Essentially: the formula z = x is absolute for transitive classes and by transitivity t x V t V so x V (and again, we could choose minimal αt On and then observe that x Vβ where β = sup {α t : t x}). Powerset: Here we show that x V z V P (x) V z: If x V, then let α On (minimal or not) such that x V α and let z = V α + 1 V so z V (and again, the utility lemma is not really needed). If t P (x) V then t V α so t V α so t z. Infinity: Exercise (go through the details which show that V = Ind(ω) by absoluteness and observe ω V ω+1 ). Foundation: Let x V and assume x. So there is α On such that x V α. Consider {β α + 1 : x V β } and note that this is a non-empty set of ordinals, so has an -minimal element µ. Note that µ cannot be a limit as x β<µ V β means that there is β < µ with x V β contradicting minimality of µ. Also µ as V =. So µ = η + 1 for some ordinal η. Pick m x V µ (which exists by construction of µ). Then m V η (by definition of V η+1 = V µ ). So if t m x then t V η x once again contradicting minimality of µ. Thus m x = as required. Theorem 9. If ZF is inconsistent, then so is ZF. Proof. Suppose there is are proofs P 1 giving ZF ψ and P 2 giving ZF ψ for a sentence ψ. Note that for every axiom φ used in the proofs above, by the previous theorem we can write down proofs of ZF φ V and follow them with P V 1 and P V 2 (i.e. where every line of P i is relativized to V ) to obtain a ZF ψ V and ZF ψ V and thus ZF is inconsistent. The precise details of course depend on your formal proof system, but it will be important that we have ZF x x V. 8.1 An alternative definition of V : There is an alternative approach to defining V. For a set x, we define the transitive closure of x as the smallest transitive set containing x and denote it by trcl(x). That a transitive set containing x exists follows from applying the Recursion Theorem (on ω) with F = x and a = x and noting that G(ω) {x} will then by transitive. We can then form the minimal one, just like the alternative definition of ω. Now, we can define a set x to be hereditarily well-founded if is well founded on trcl(x) and let V = {x : x is hereditarily well-founded}. We can then show that is well founded on every subset of V (so V = Foundation) and also that applying any axiom of ZF to hereditarily well-founded sets gives new sets 20

21 which are hereditarily well-founded. This will then (with a bit of extra work) show that V = ZF. We can recover the V α by observing that in fact is well-founded on V and recursively defining V α = x V \ V β : x is -minimal in V \ β<α although it is not clear that the V α thus defined are sets! See Kunen for details on this approach. It is a nice approach in that we explicitly construct the largest V which could be well-founded (and transitive) and then check that it works. Our approach is to discover V and it is pure luck that it does satisfy ZF. 9 Gödel s Constructible Universe L We will now work in ZF (and note the instances of Powerset we will use) to define a smaller universe L. Our surrounding universe is now V = {x : x = x} (under ZF this is the same as V U as defined in the previous section, but I want to emphasize that we assume Foundation). 9.1 The Definable Subsets We want to define { there is a formula φ(v Def(x) = y : 1,..., v n, v n+1 ) and a 1,..., a n x such that t [t y t x (x, ) = φ(a 1,..., a n, t)] Of course, we cannot do this formally since we cannot quantify over formulae! However, there is a workaround. We can internalize (x, ) = φ(a 1,..., a n ) within any class A which satisfies enough of ZF Powerset. More precisely there is a set F A (in fact F ω), a class function val: A ω A {0, 1} and a function free: ω ω <ω, free A, such that whenever φ(v k1,..., v kn ) is a formula of LST with all free variables shown then there is φ F such that ZF Powerset {k 1,..., k n } = free( φ ) ZF Powerset x A a x <ω [val(x, φ, a) = 1 [free( φ ) dom(a) φ(a(k 1 ),..., a(k n )) x ]]. Using this, we can define y Def(x) m F a x free(m)\{1} [1 free(m) t x [t y val(x, m, a { 1, t }) = 1]]. For later, we note that if A B are non-empty transitive classes satisfying enough of ZF Powerset then F as well as val and hence y Def(x) are absolute for A, B. β<α V β }. 21

22 We can also see that y Def(x) y x so that if A satisfies additionally Powerset (and a further instance of Separation) then in fact the above defines the class function Def : A A via Def(x) = {y P (x) : y Def(x)}. In fact, there is no need for Powerset: you can use Replacement instead The details of defining Def - Non-Examinable There are a lot of different ways of defining val, F and free as well as φ. We present one of them, the details of which are not important. We first note that we can define the usual arithmetic functions on ω (interpreted as N) by the Recursion Theorem and that these are absolute. In the meta-theory, we then use a nice Gödel numbering of the formulae of LST (although this is only relevant at the very end - but it does help understanding). Of course this does depend on our language, so we need to fix it: The terms are v... (or more formally we define recursively t 0 = {v }, t n+1 = {s : s t n }) and we code them by t = { 2; t = v 2 s ; t = s So terms are powers of 2 and we write v k instead of v... (k s) (for sanity reasons) and we let T = { 2 k : k ω, k 1 }. Next, the atomic formulae are (for t, s terms, so we can think t, s T ) coded by t = s t s t = s =3 t 5 s 7 1 t s =3 t 5 s 7 2 and we let A = { 3 t 5 s 7 k : t, s T, k {1, 2} }. Finally, the formulae are coded by φ; φ an atomic formula φ; φ a formula φ ψ; ψ, φ formulae v k φ; v k a term, φ a formula φ =3 φ 7 3 φ ψ =3 φ 5 ψ 7 4 v k φ =3 φ 5 vk

23 and we let F = A { 3 p 7 3 : p F } { 3 p 5 q 7 4 : p, q F } { 3 p 5 t 7 5 : p F, t T }. Of course, the definition of F doesn t seem to make sense, so we should (by recursion on ω) set F 0 =A F n+1 =F 0 { 3 p 7 3 : p F n } { 3 p 5 q 7 4 : p, q F n } { 3 p 5 t 7 5 : p F n, t T } F = n ω F n. For convenience, we have chosen a minimal language. It is not difficult to see how to deal with a more complicated language having all logical connectives as well as existential quantifiers. We note that T, A, F can be defined in a sufficiently large fragment of ZF Powerset and are absolute for transitive non-empty models of this fragment. Now we define the function free on ω which takes values in ω <ω (finite functions into ω, formally I think I want ω <ω = {f : F ω : F finite ω}) as follows (by recursion on ω): free(0) = {0} {0} ; n + 1 F free(k); n + 1 F n + 1 = 3 k 7 3 free(n + 1) = free(k) free(l); n + 1 F n + 1 = 3 k 5 l 7 4 free(k) \ {l} ; n + 1 F n + 1 = 3 k 5 l 7 5. You should convince yourself that free gives {0} if the input is not (the code for) a formula and otherwise the set of free variables in the formula. (Note that I have made sure that 0 T so that 0 free(k) if and only if k F.) We observe that free is absolute for non-empty transitive classes satisfying enough of ZF Powerset. Finally, given x, we can define a function val x : ω x <ω {0, 1, 2} by recursion on ω (here I interpret x <ω = {a : b x : b finite ω}). Essentially, this is the model-theoretic satisfaction relation = adapted to our representation of φ by φ. In the big case distinction, I have put the logical symbol which is 23

24 dealt with in square brackets before the n + 1. val x (0, a) = 0 0; n + 1 F 0; n + 1 F free(n + 1) dom(a) [ ] 0; [=] n + 1 F free(n + 1) dom(a) k, l ω n + 1 = 3 2k 5 2l 7 1 a(k) a(l) [ ] 1; [=] n + 1 F free(n + 1) dom(a) k, l ω n + 1 = 3 2k 5 2l 7 1 a(k) = a(l) [ ] 0; [ ] n + 1 F free(n + 1) dom(a) k, l ω n + 1 = 3 2k 5 2l 7 2 a(k) a(l) [ ] 1; [ ] n + 1 F free(n + 1) dom(a) k, l ω n + 1 = 3 2k 5 2l 7 2 a(k) a(l) 0; [ ] n + 1 F free(n + 1) dom(a) k, l ω [ n + 1 = 3 k 7 3 val x (k, a) = 1 ] 1; [ ] n + 1 F free(n + 1) dom(a) k, l ω [ n + 1 = 3 k 7 3 val x (k, a) = 0 ] val x (n + 1, a) = 0; [ ] n + 1 F free(n + 1) dom(a) k, l ω [ n + 1 = 3 k 5 l 7 4 [val x (k, a) = 0 val x (l, a) = 0]] 1; [ ] n + 1 F free(n + 1) dom(a) k, l ω [ n + 1 = 3 k 5 l 7 4 [val x (k, a) = 1 val x (l, a) = 1]] 0; [ ] n + 1 F free(n + 1) dom(a) k, l ω [ n + 1 = 3 k 5 l 7 5 â x [ <ω â free(n+1)\{l} = a free(n+1)\{l} l dom(â) val x (k, â) = 0 ]] 1; [ ] n + 1 F free(n + 1) dom(a) k, l ω [ n + 1 = 3 k 5 l 7 5 â x [ <ω â free(n+1)\{l} = a free(n+1)\{l} l dom(â) val x (k, â) = 1 ]] 0 otherwise - this case should not occur Note that because x <ω is absolute (for transitive non-empty classes satisfying enough of ZF Powerset), val x is in fact absolute for these transitive non-empty classes. For a formula φ(v k1,..., v kn ) of LST with all free variables shown, and a 1,..., a n x we define (x, ) = φ(a 1,..., a n ) val x ( φ, { k i, a i : i = 1,..., n}) = A few definable sets We work in any non-empty transitive class A satisfying enough of ZF Powerset. Lemma 9. Suppose x is a set. 1. Def( ) hence Def( ) = { }. 2. x Def(x). 3. For y, z x, {y, z} Def(x). More generally, finite subsets of x are definable. Proof. We give the relevant formulae and parameters and leave it to the reader to verify that these define the appropriate sets: 24

25 1. φ(v 1 ) v 1 v 1 and a = ; 2. φ(v 1 ) v 1 = v 1 and a = 3. φ(v 1, v 2, v 3 ) v 1 = v 3 v 2 = v 3 and a = { 1, y, 2, z }; in general, we use φ(v 1,..., v n, v n+1 ) v 1 = v n+1 v n = v n+1 to obtain sets with (up to) n elements; Remark 6. We will in fact spell out what really happens in the last of these cases (when we take A = V ), just to convince you that you really don t want to do this (without a computer), but that it is not hard, just tedious. Note that since val is absolute for non-empty transitive classes satisfying enough of ZF Powerset we get that (x, ) = φ(a 1,..., a n ) is absolute for these classes. Also φ x φ φ A, so taking A = V does not lose generality. First we work out v 1 = v 3 v 2 = v 3. Note that for use, was only an abbreviation so v 1 = v 3 v 2 = v 3 [ [v 1 = v 3 ] [v 2 = v 3 ]]. With our concrete definition, n 1 = v 1 = v 3 = and Then and Thus and finally n 2 = v 2 = v 3 = n 3 = [v 1 = v 3 ] = 3 n1 7 3 n 4 = [v 2 = v 3 ] = 3 n n 5 = [v 1 = v 3 ] [v 2 = v 3 ] = 3 n3 5 n n = [ [v 1 = v 3 ] [v 2 = v 3 ]] = 3 n = Clearly (well, by construction) n F and free(n) = {1, 2, 3}. We now need to check that t {y, z} val (x, n, a { 3, t }) = 1. So suppose that t = y. So, let us work out val (x, n 1, a { 3, t }) = 1 since n 1 F, free( ) = {1, 3} {1, 2, 3} = dom (a { 3, t }) and k = 1, l3 are such that n 1 = 3 2k 5 2l 7 and in fact a(1) = y = t = a(3). Thus val (x, n 3, a { 3, t }) = 0 25

26 and so giving val (x, n 5, a { 3, t }) = 0 val (x, n, a { 3, t }) = 1 as required (we left out the detailed checks for these last three claims). The case t = z works similarly. For the converse, if t y and t z then val (x, n 1, a { 3, t }) = 0 as well as val (x, n 2, a { 3, t }) = 0. Thus val (x, n 3, a { 3, t }) = 1 as well as val (x, n 3, a { 3, t }) = 1 giving val (x, n 5, a { 3, t }) = 1 so that val x, n, a { 3, t } = The definable sets L Definition 10. By recursion on On we define a class function L : On V (strictly L V since our definition happens relativized to V ) L 0 = ; α On L α+1 = Def (L α ); γ Lim L γ = β<γ L β We also write L for the class α On L α Lemma 10. ZF proves: Forall α On: 1. L α V α ; 2. L α L α+1 ; 3. L α is transitive and L α L α+1, 4. L α On = α and α L α+1 Proof. 1. By induction on α: base case and limit stages are trivial; for successor steps note that Def (x) P (x). 2. See Lemma By induction on α: again, the base case and the limit stages are straightforward. So suppose x L α+1. As L α+1 is transitive x = {t : t L α t x} Def (L α+1 ) = L α+2 giving. Now assume that t x L α+2. Then x L α+1 so t L α+1 L α+2. 26

27 4. Again, by induction on α: The non-trivial bit of the base case has been covered in Lemma 9. For the successor step: assume L α On = α and α L α+1. Let β L α+1 On. If α β L α then α L α On = α, a contradiction. So β = α or β α (as totally orders On) and hence β α {α} = α + 1. Conversely, assume that β α + 1. Either β = α and then by inductive assumption β = α L α+1 or β α L α+1 and by transitivity of L α+1 we obtain β L α+1. Also φ(v 1, v 2 ) v 1 v 2 v 1 = v 2 applied with v 2 = α shows that α + 1 L α+2. For the limit: suppose γ Lim and the result is true for α < γ. If β L γ On then β L α for some α < γ and hence β γ. Conversely, if β γ then β < γ so β L β+1 L γ. But then φ(v 1 ) v 1 is an ordinal shows that γ Def (L γ ) as required. 9.4 ZF in L I Theorem 10. For each φ ZF, ZF φ L i.e. L = ZF. Proof of φ L for the easy axioms φ. We check the axioms in turn. Extensionality: L is transitive, so satisfies Extensionality L (like V does). Emptyset: V L and z = is absolute, so L = V witnesses Emptyset L. Pairing: Suppose x, y L and find α On such that x, y L α+1. By an earlier lemma {x, y} V Def (L α+1 ) = L α+2 L and by absoluteness of z = {x, y}, {x, y} L = {x, y} V witnesses Pairing L (for x, y). Union: Suppose x L and find α On such that x L α+1. Let z = ( x) V. Note that if t y x L α+1 then by transitivity of L α+1 we have t L α+1 so that z L α+1. By definition of z, t z [t L α+1 y x [t y]]. The formula y x [t y] is 0 so absolute and hence ] t z [t L α+1 [ y x [t y]] Lα+1. But then t z [t L α+1 (L α+1, ) = y x [t y])]. 27

28 Thus we let φ(v 1, v 2 ) v 3 v 1 [v 2 v 3 ] (and a 1 = x) to see that z DefL α+1 = L α+2. Finally, by absoluteness of z = x, z = ( x) L witnesses Union L (for x). Powerset: Suppose x L and find α x On such that x L αx. Let z = P (x) V L, so that t z t x t L. For each t z, let α t On be minimal such that t L αt+1 and let α = sup {α t + 1 : t z} {α x } On so that z L α and x L α. Hence by absoluteness of t x. Thus t z t x t L α [t x] Lα t L α t z t L α (L α, ) = t x and hence φ(v 1, v 2 ) v 2 v 1 v 3 v 2 [v 3 v 1 ] (and a 1 = x) witnesses that z Def(L α ) = L α+1 L. We now need observe that the above discussion shows [ t [t z t x]] L t L [t z [t x] L] so that indeed z = P (x) L witnesses Powerset L (for x). Foundation: If x L then x V so find m V such that [m is -minimal in x] V. Then m x L gives m L and being -minimal in x is absolute ( y x [y m] is 0 ), so [m is -minimal in x] L as required. Infinity: By the Lemma ω V L ω+1 and φ(z) z x z [x {x} z] is absolute. Since φ(ω V ) V we obtain φ(ω V ) L as required (and we could note that ω L = ω V ). Remark 7. It is worth noting that the above proof shows that Each L α satsifies Extensionality and Foundation and if ω α then L α satisfies Infinity. For every limit ordinal γ, L γ satisfies Pairing and Union. Proof attempt at instances of Separation L. Suppose φ(v 1,..., v n, v n+1 ) is a formula of LST with all free variables displayed and that a 1,..., a n, x L. As before we find α L such that a 1,..., a n, x L α. By Separation in V applied with the formula φ L, we obtain z = { t x : φ(a 1,..., a n, t) L} and since z x L α we obtain z L α. 28

29 So we may be tempted to show that z DefL α with the formula φ(v 1,..., v n, v n+1 ) L t x. This encounters a problem: first the relativization L has to be relativized itself. Let s write ψ L (v[ n+2 ) for the formula defining L. If for example φ v n+2 ψ then φ L v n+2 ψl (v n+2 ) ψ L], so that ( φ L ) L α vn+2 [v n+2 L α ψ L (v n+2 ) Lα ( ψ L) ] L α. At this stage we are in no position to prove that ψ L is absolute for L α, V because L α might not satisfy all of ZF we need (even if we assume that α is a limit ordinal). So, to avoid this, we try again, this time maybe with φ(v 1,..., v n, v n+1 ) t x since we automatically relativize to L α+1 when checking considering (L α+1, ) = φ(a 1,..., a n, t). But here we also encounter a problem. To carry out this plan, we will need to show that and we only know that t z t x φ(a 1,..., a n, t) Lα t z t x φ(a 1,..., a n, t) L. Thus we need to get our hands on some ordinal γ α such that φ(v 1,..., v n, v n+1 ) is absolute for L γ, L. Provided we have this γ, we then have t z t x φ(a 1,..., a n, t) Lγ so that z Def (L γ ). This γ does exist as we will show in the next theorem, the Levy Reflection Principle. Then by consruction [z = {t x : φ(a 1,..., a n, t)}] L and we are done. The Levy Reflection Principle We aim to show the following meta-theorem. First the informal, wordy version: Theorem 11. For every formula φ of LST, if A : On V is a non-decreasing class function of non-empty transitive sets and B = α On A α then there are arbitrarily large limit ordinals γ such that φ is absolute for A γ, B. Or, more formally. Theorem 12. Suppose A is a formula of LST with one free variable and that φ(v 1,..., v n ) is some formula of LST. We will assume that A is a class function on On (see below) and then write A α = A(α) and B = α On A α = {x : α On} x A α ZF Powerset proves: A is a class function on On A 1 α, β On [α β A α A β ] α On A α is transitive α On γ Lim [α γ φ is absolute for A α, B]. 29

30 where of course φ is absolute for A α, B means a 1,..., a n A α [ φ(a1,..., a n ) Aα φ(a 1,..., a n ) B] We first prove a lemma, the well known Tarski Vaught criterion for absoluteness (from Model Theory). But first we need a definition in the meta-theory. Definition 11. A list φ 1,..., φ n of formulae of LST is subformula closed if and only if every subformula of each φ k appears in the list. Note that whenever φ 1,..., φ n is a list of formulae of LST, we can add all the subformulae to obtain a subformula closed list φ 1,..., φ n,..., φ m. Lemma 11 (Tarksi Vaught criterion). Suppose φ 1,..., φ n is a subformula closed list of formulae of LST which do not contain the universal quantifier (but may contain ) and that A B are transitive, non-empty classes. Then each of φ 1,..., φ n is absolute for A, B provided that for every formula φ k (v 1,..., v m ) of the form v m+1 φ j (v 1,..., v n, v m+1 ) we have a 1,..., a m A [ φ(a 1,..., a m ) B a m+1 A φ(a 1,..., a n, a m+1 ) B]. Proof. By induction on the complexity of the (most complicated) formula. Atomic formulae are trivially absolute. The logical connectives,,,, are trivial by induction. Thus assume that the most complicated formula is φ n (v 1,..., v m ) v m+1 φ j (v 1,..., v m, v m+1 ) and that the statement is true for simpler formulae. In particular, φ j is absolute for A, B. So let a 1,..., a m A. Now observe φ(a 1,..., a m ) A a m+1 A φ j (a 1,..., a m, a m+1 ) A. Since φ j is absolute [ am+1 A φ j (a 1,..., a m, a m+1 ) A] [ a m+1 A φ j (a 1,..., a m, a m+1 ) B]. Since A B (for the direction) and by the assumption (for the -direction) [ am+1 A φ j (a 1,..., a m, a m+1 ) B] φ(a 1,..., a m ) B. Hence φ(a 1,..., a m ) A φ(a 1,..., a m ) B as required. The Tarski Vaught criterion can be understood as a closure criterion: if an existential formula φ is true in B then we can in fact find an element of A witnessing φ B. To prove the Levy Reflection principle, we emulate the proof of the Skolem Theorem, constructing the Skolem closure. We start with some A α and add witnesses to all existential subformulae in B that are true in B. Of course, adding these witnesses gives us new parameters so that we have to consider more instantiations of formulae. As a concrete example, consider (in the language of fields) the statement φ(x) y y 2 = x. Then φ is not absolute for Q, C (because 2 Q and 30